When \(\Gamma ={\mathrm {Cay}}(\mathbb {Z}_{2}\times \mathbb {Z}_{2}, \Phi )\) , sigraph \(\Sigma \) 
has two connected components. One of the components has a positive edge and the other
component has a negative edge. (According to the explanation given in the last paragraph of the
introduction, \(V(\Sigma _{1})=\lbrace (0, 0), (1, 1)\rbrace \)  and \(V(\Sigma _{2})=\lbrace (0, 1), (1, 0)\rbrace \) . Since
\((1, 1)\in \Phi \)  and \((0, 0)\)  is adjacent to \((1, 1)\) , so, by definition of \(\sigma \) , the edge \((0, 0)(1, 1)\) 
is positive. Also, since the vertices of \(V(\Sigma _{2})\)  are not in \(\Phi \)  and \((0, 1)\)  is adjacent to \((1, 0)\) ,
hence \((0, 1)(1, 0)\)  is negative edge). Note that the tree has no cycle and using Theorem 1.6 [1]},
we can conclude that each tree is balanced. Since \(\Sigma \)  is the union of two trees, so it is
balanced. Also \(L(\Sigma )\)  has only two vertices so we may consider it is a balansed graph. Therefore
in the proof of Theorems REF , REF , this case is not considered.
